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Math Help - Fourier series and (possibly) Parsevals Theorem

  1. #1
    Super Member Deadstar's Avatar
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    Fourier series and (possibly) Parsevals Theorem

    Let f be the function given by f(\theta) = |\theta| for \theta\in [ -\pi , \pi ].

    a) Calculate the fourier coefficients of f and show that...

    \hat{f}(n) = \frac{(-1)^n - 1}{\pi n^2} for n \neq 0

    \hat{f}(n) = \frac{\pi}{2} for n = 0.

    done...

    b) What is the Fourier series of f in terms of sines and cosines?

    My answer, \frac{\pi}{2} + \sum_{n=-\infty}^{\infty} \frac{(-1)^n - 1}{\pi n^2} \cos(n\theta).

    c) Taking \theta= 0, prove that

    \sum_{k=0}^{\infty} \frac{1}{(2k+1)^2} = \frac{\pi^2}{8} and \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}

    Now this, I suspect, should have just been a simple application of Parsevals theorem but I'm going wrong somewhere.

    Parsevals Theorem.

    \left ( \frac{1}{2 \pi} \int_{-\pi}^{\pi} |f(\theta)|^2 d \theta \right)^{\frac{1}{2}} = \left ( \sum_{n=-\infty}^{\infty} |\hat{f}(n)|^2\right)^{\frac{1}{2}}

    But with \theta = 0 the left hand side is 0.

    Thus we get,

    0 = \left ( \sum_{n=-\infty}^{\infty} |\hat{f}(n)|^2\right)^{\frac{1}{2}}

    = \left ( \frac{\pi^2}{4} + \sum_{n=-\infty}^{\infty} \left (\frac{1-1}{n^2 \pi} \right )^2 \right)^{\frac{1}{2}} = \frac{\pi}{2}

    So clearly I'm doing something wrong... Most likely in that last 2 steps...
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  2. #2
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    Quote Originally Posted by Deadstar View Post
    Let f be the function given by f(\theta) = |\theta| for \theta\in [ -\pi , \pi ].

    a) Calculate the fourier coefficients of f and show that...

    \hat{f}(n) = \frac{(-1)^n - 1}{\pi n^2} for n \neq 0

    \hat{f}(n) = \frac{\pi}{2} for n = 0.

    done...

    b) What is the Fourier series of f in terms of sines and cosines?

    My answer, \frac{\pi}{2} + \sum_{n=-\infty}^{\infty} \frac{(-1)^n - 1}{\pi n^2} \cos(n\theta).


    For even n's the summands of this series are zero, so you can take only odd n's in the sum and since you have a square in the denominator this

    is the same as twice the sum over the odd natural numbers (taking \theta=0):

    \frac{\pi}{2} + \sum_{n=-\infty}^{\infty} \frac{(-1)^n - 1}{\pi n^2} \cos(n\cdot 0)=\frac{\pi}{2}+\frac{2}{\pi}\sum^\infty_{k=0}\fr  ac{-2}{(2k-1)^2} = \frac{\pi}{2}-\frac{4}{\pi}\sum^\infty_{k=0}\frac{1}{(2k-1)^2} .

    Now equal the above to zero and you'll get what you want...

    Tonio



    c) Taking \theta= 0, prove that

    \sum_{k=0}^{\infty} \frac{1}{(2k+1)^2} = \frac{\pi^2}{8} and \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}

    Now this, I suspect, should have just been a simple application of Parsevals theorem but I'm going wrong somewhere.

    Parsevals Theorem.

    \left ( \frac{1}{2 \pi} \int_{-\pi}^{\pi} |f(\theta)|^2 d \theta \right)^{\frac{1}{2}} = \left ( \sum_{n=-\infty}^{\infty} |\hat{f}(n)|^2\right)^{\frac{1}{2}}

    But with \theta = 0 the left hand side is 0.

    Thus we get,

    0 = \left ( \sum_{n=-\infty}^{\infty} |\hat{f}(n)|^2\right)^{\frac{1}{2}}

    = \left ( \frac{\pi^2}{4} + \sum_{n=-\infty}^{\infty} \left (\frac{1-1}{n^2 \pi} \right )^2 \right)^{\frac{1}{2}} = \frac{\pi}{2}

    So clearly I'm doing something wrong... Most likely in that last 2 steps...
    .
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